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NCERT solutions for class 9 Science chapter 8 Motion

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NCERT solutions for class 9 Science


Motion


8.1 Describing motion

Change in the position of an object with respect to time is called motion.
Example –

  • The motion of a car
  • Birds flying
  • The motion of planets stars and galaxies.

8.1.1-Reference point-

The point from which motion of an object can be observed. It is also called the origin.
Example- The shopping mall is 3km south of my home. Here, we are specifying the position of the mall with respect to my home so my home is a reference point in this example.

8.1.2 -Motion along a straight line-

•This is the simplest type of motion. It can be explained as follow.
•Let us assume a person starts his journey from point O that is its origin. He moves towards his destination “A”.Again follow the same path and reaches the next destination “ C”.

•Distance –

The total path traveled by an object.
Scalar quantity means it only shows magnitude, not direction.
It cannot be zero.
Total path travelled= 0A+0C
=60km+35km
=95km

•Displacement-

The shortest distance measured between the final position and the initial position of an object.
– It is a vector quantity that means it has magnitude as well as direction.
– It can be zero
– Displacement of an object from 0 to B is 35 km in the above example.

8.1.3- Uniform motion and non-uniform motion

• Uniform motion –

When an object covers equal distance in an equal time interval.
Example- A bus driver covers 60 km in the first hour and then again 60kilometre in the second hour.

•Non- uniform motion-

An object covers the unequal distance in an equal time interval.
Example- A bus driver covers 65 km in the first hour and 70 kilometres in second hours. This is an example of non-uniform motion.


8.2-Measuring the rate of motion-

• Speed –

An object in unit time is called speed.SI unit of speed is metre per second, centimetre per second, and kilometre per hour To find the speed of an object we require magnitude and it is not
necessary that the speed of an object be always constant.
It is a scalar quantity
Speed=Distance /time

For Non -uniform motion we need to find the average speed that is a ratio of total distance travel and total time is taken.
Average speed= Total distance travelled/total time taken

•Velocity-

Velocity is the speed of definite direction.
It can be constant or variable.
Its unit is meter per second (m/s).
It is a vector quantity
When the velocity of an object change with uniform rate then, in that case, we find the average velocity
that is arithmetic mean of initial velocity and final velocity
Average velocity= initial velocity + final velocity / 2
V [av] =u+v /2


8.3- Rate of change of velocity-

Acceleration-

It is defined as a rate of a change in the velocity of an object per unit time
Acceleration = change in velocity/ time taken
a = v – u / t
The value of “a “ is positive when the direction of velocity is positive a negative is when it is a positive direction of velocity.
Unit -meter per second square (m/s²)
The object is said to be in a uniform accelerated when that object travels in a straight line and its velocity increases or decreases by equal time interval whereas it is said to be in non-uniform acceleration when its velocity changes with a non- uniform rate.


8.4- Graphical representation of motion

The graph is a convenient method to present basic information about a variety of motion. Some of the graph representations are as follow:

8.4.1-Distance -time graph:

1- Distance -time graph for uniform motion-

Time is taken along the x-axis and distances taken on the y-axis
Let us consider the small part AB, the velocity along this path is given by
v= s2-s1 / t2-t1

2-Distance -time graph for non-uniform motion-

8.4.2- Velocity -Time graphs

The variation in the velocity with time for an object moving in a straight line can be represented by the uniform velocity-time graph in this the graph the time it presented in the x-axis whereas the velocity represented in the y-axis.

1- Velocity- time graph for the uniform motion of a car

Distance travelled by car in time t2-t1 can be expressed as an area of rectangle ABDC.

2- Velocity time graph for a car moving with a uniform acceleration

S= area ABCDE.
= area of rectangle ABCD +area of triangle ADE


8.5-Equation of motion by graphical method-

Equations of motion are-

V=u+at ………..(i)

S=ut+½at² …….(ii)
V²=u² +2as ……..(iii)
Where

u-initial velocity
v-final velocity
a- acceleration
t-time
s-distance

8.5.1- Equation for velocity-time relation

This is a velocity-time graph of an object moving with uniform acceleration.
point A=u

point B=v
Change in velocity at uniform rate =a
Draw perpendicular from Point B to the x-axis (point E)and y-axis (point C)
so that,
Initial velocity=OA
Final velocity=BC
Time interval =O
Now,BD=BC-CD

Draw AD parallel to OC
BC=BD+DC=BD+OA
V=BD+u
BD=v-u—————-(a)
Now,
a=BD/AD=BD/OC=BD/t
BD=at——————(b)

Put value of equation (b) in equation (a) we get
at =v-u
v= u+at (This is the first equation of motion)

8.5.2-Equation for position-time relation

Distance travelled by object = S
time taken =t
acceleration = a
Now, according to the above velocity-time graph
. Distance travelled by an object =area of OABC
= area of rectangle OADC + area of triangle ABD
= OA x OC+½(ADxBD)
=ut+½(t x at)
S=ut+½ at² (This is the second equation of motion)

8.5.3- Equation for position -velocity relation

Distance travelled=S
time taken=t
Uniform acceleration = a.
S=area of trapezium O ABC
=(OA +BC) X OC / 2
=(u+v ) t/2………..(i)
And we know t = (v -u)/a…………(ii)
Put equation (ii) in equation (i)
We get, s= ( v+u)x( v-u)/2a
2as = v2-u2 This is third equation of motion


8.6 – Circular motion –

The motion of an object in a circular path or along a circumference of a circle. In circular motion, initial position and final position are at the same point or we can say displacement is zero.

•Uniform circular motion –

When an object moves in a circular path with uniform speed, its motion is called uniform circular motion.
As we are talking about the circular path, that means motion along the circumference of the circle of radius ‘r’ is given by =2πr
Time is taken by the object to move in circular path =t
Speed (v)= 2πr/t
•When an object moves in a circular motion, its velocity changes only due to change in direction, not magnitude.
Example- Motion of satellite around the earth.


NCERT Solutions


Page: 100


1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer-Yes, an object e can have zero total displacement. As we can define displacement refers to the shortest distance between the initial and the final positions of the object. If an object starts and ends with the same point then a difference of initial and final position becomes zero and hence displacement would be zero.

2. A farmer moves along the boundary of a square field of side 10m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Answer- Total distance travelled =4 X side
=4 X 10 = 40 m

Time taken =40 sec
Distance travelled in 1 sec= 40m/40 sec =1m
Total distance travelled by farmer in 2min 20 sec [140 sec]=140 m
Now, the total displacement of the farmer depends on the initial position.
the total displacement of the farmer will be equal to the length of the diagonal of a square field

Pythagoras theorem,
the length of the diagonal
√(10²+10²)
= √200= 14.14m.
This is the maximum possible displacement of the farmer.

3. Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.

Answer- (a)False
(b)False


Page: 102


1. Distinguish between speed and velocity.
Answer- 
Velocity –

  • It is defined as the ratio of displacement and time taken.
  • It is a vector quantity.
  • Velocity = displacement/time.
  • Velocity can be zero

Speed-

  • It is defined as the ratio of distance and time.
  • It Is scalar quantity.
  • Speed = distance / time
  •  Speed cannot have zero value.

2. Under what condition(s) is the magnitude of the average velocity of an object equal to its average speed?
Answer-The magnitude of the average velocity of an object is equal to its average speed is can possible in the condition only when the Total distance travelled by an object is equal to the total displacement of an object.

3. What does the odometer of an automobile measure?
Answer-The odometer measures the total distance travelled by automobile.

4. What does the path of an object look like when it is in uniform motion?
Answer-The path of an object in uniform motion is a straight line.

5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 m/s.

Answer-Time is taken by the signal to reach the ground station –

5 minutes = 5 X 60 seconds = 300 seconds.
The speed of the signal is given as = 3 × 108  m/s.
Therefore, total distance = (3 × 108  m/s) X 300s
= 9X1010 meters.


Page: 103


1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration? 

Answer-

(i) Uniform Acceleration: When that object travel in a straight line and its velocity increases or decreases by equal time interval.
(ii) Non-uniform acceleration: When its velocity changes with a non- uniform rate.

2. A bus decreases its speed from 80 km h–1 to 60 km h–1 in 5 s. Find the acceleration of the bus.
Answer- Initial velocity (u)= 80km/hour = 22.22 m/s
final velocity (v)= 60km/hour = 16.66 m/s
Time (t) = 5 seconds.
Acceleration (a)=(v-u)/t
a= (16.66 m/s – 22.22 m/s)/5s
a= -1.112 m/s²


3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h–1 in 10 minutes. Find its acceleration.

Answer- u of the train = 0m/s1
v of the train = 40km/hour = 11.11 m/s1
Time t = 10 minutes = 600 s.
acceleration a=(v-u)/t
= (11.11 m/s – 0 m/s)/600s
= 0.0185 m/s²


Page:107


1-What is the nature of the distance-time graphs for uniform and non- uniform motion of an object?

Answer –Uniform motion- The slope is a straight line.
Non-uniform motion –The slope is a curved line.

2-. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer-This says that the object is at rest.

3. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Answer-Object is said to be in uniform motion.

4. What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer-Displacement is measured by the area occupied below the velocity-time graph.


Page: 109,110


1. A bus starting from rest moves with a uniform acceleration of 0.1 m/s² for 2 minutes. Find

(a) the speed acquired, (b) the distance travelled.
Answer- (a) Initial velocity (u) = 0 m/s As bus starts from rest
Acceleration (a) = 0.1 m/s²
Time (t) = 2 minutes
2 x60
= 120 s
According to first equation

v= u +at
= 0 m/s
1+(0.1 m/s
2 X 120s)

= 12m/s+ 0 m/.s
Therefore, (v) = 12m/s
(b) third motion equation,
2as =v² – u²

=2 x 0.1 x s =(12² – 0²)

s = 720m
The total distance traveled is 720m.

2. A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of –0.5 m/s2.Find how far the train will go before it is brought to rest.

Answer-U = 90 km/hour =90x 5/18 = 25 m/s
v= 0 m.s-1
Acceleration a = -0.5 m.s-2
A/c to third motion equation,
v² – u² =2as

distance
(s) =(v2-u2)/2a

s = (02-252)/2(-0.5) meters = 625 meters

Distance travelled=625 m

3. A trolley, while going down an inclined plane, has an acceleration of 2 cm/s² . What will be its velocity 3 s after the start?
Answer- Initial velocity =u= 0
Acceleration a= 0.02 m/s²
Time t= 3s
According to first motion equation,
v=u+at
Therefore, v= 0 + (0.02 m/s
2 X 3s)
= 0.06 m/s²

4. A racing car has a uniform acceleration of 4 m/s² . What distance will it cover in 10 s after start?
Answer- Initial velocity (u) = 0 m/s
Acceleration (a) = 4 m/s²
Time period (t) = 10 s
According to second motion equation,
s = ut+½ at²
(s) = 0 x10m + ½ x4m/s²X10s²
= 200 meters

5. A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m/s² in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer-Initial velocity = 5 m/s =u
Final velocity = 0 m/s=v
Acceleration =- 10 ms-2 = as it is in direction opposite to the trajectory of the stone

According to third equation of motion
v² – u² =2as

Therefore, the distance traveled by the s.
(s) = (02 – 52)/ 2(10)
Distance (s) = 1.25 meters
According first motion equation, v = u + at
t= (v – u) /a
t=(0-5)/-10 s
Time taken = 0.5 seconds


Exercise


1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Answer-
Diameter of circular track = 200m

The distance of circular field =circumference of the track (2πr) = 200π meters.
Time taken is 40 seconds to cover 200π meters
Distance covered in 1 second = 200π/40
If 2minutes and 20 seconds =140 seconds
Distance covered in 140 sec
= 140 X200π/40 meters
= 140X200X3.14/40 meters
Total distance = 2200 m
Net displacement= diameter =200 m

2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Answer- Distance covered from point A to B = 300 meters
Time is taken to travel from point A to B = 2 minutes and 30 seconds or 150 seconds
Displacement from A to B = 300 -0=300 m

Distance covered from point A to C = AB+BC
300m + 100m = 400 meters
Time is taken to travel from point A to C = time taken from A to B +B to C
=2 min 30 secs + 1 min = 210 seconds
Displacement from A to C = 300m – 100m = 200 m

A)the average speed while traveling from A to B = 300/150 m/s = 2 m/s
Average velocity while traveling from A to B =300/150 m/s = 2 m/s

B)Average speed while traveling from A to C = 400/210 m/s
-= 1.9 m/s
Average velocity while traveling from A to C =200/210 m/s= 0.95 m/s

3. Abdul, while driving to school, computes the average speed for his trip to be 20 km/h.On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed for Abdul’s trip?

Answer-Let we assume, Distance traveled to reach the school = distance traveled to reach home = x
And Time is taken to reach school = t1
Time is taken to reach home = t2
, Therefore, t1 = d/20 and t2 = d/30
Now, the average speed for the entire trip is given by total distance travelled/ total time taken
= (x +x )/(t1+t2)Km/h
= (x +x )/(x/20+x/30)Km/h
=120x /5x
= 120/5 km/h
= 24 km/h

4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s² for 8.0 s. How far does the boat travel during this time?
Answer-According to second motion equation, s = ut + ½at²
Where u = 0 m/s
Acceleration a = 3 m/s²
Time period [t]= 8s
Put values in above equation we get
distance travelled by boat s= 0 X 8 + 1/2 x3 x (8)²
=0+96 m
= 96 m

5. A driver of a car traveling at 52 km h–1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

Answer-For car A ,velocity is 52 km /h [14.44 m/s]
Time is 5 sec
That means on plotting the graph
OB = 5 seconds
OA= 14.44 m/s
For car B ,velocity is 3km /h [0.83 m /s]
Time = 10 sec
That means
OD = 10 seconds
OC = 0.83 m/s

Plot of speed versus time graph for the two cars is:

Displacement of car A is given by area of triangle AOB =1/2. XOB X OA

=(1/2) X(5s)X(14.44m/s)

= 36 m

displacement of the car B is given by the area of the triangle COD
= (1/2)X(OD) X (OC)
= (1/2) X (10s)X(0.83m/s)

= 4.15 m
Therefore, the first car travelled farther

6. Fig 8.11 shows the distance-time graph of three objects A, B, and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C?

Answer-(a) B, is traveling at the fastest speed as the slope of B is the greater

(b), No three of them are never at the same point on the road As slopes are not intersecting each other.
(c) 1 graph unit =4/7 km.
Initial distance from the origin is 4 X 4/7 km = 16/7 km

Distance traveled by C = 8 – (16/7) km= 5.71 km
(d)Therefore, total distance traveled by B when it crosses C = 9 X4/7 = 5.14 km

7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s², with what velocity will it strike the ground? After what time will it strike the ground?
Answer-It is given in the question
Initial velocity of the ball = 0
Acceleration = 10 m/s2
Distance = 20m
According to third equation of motion
V²=u² +2as
V²= 2 X 10m/s2 X 20m + 0
V² = 400
v= 20m/s that is final velocity

Now according to the first equation, of motion
v =u +at
20= 0+ 10t
= 2 seconds
The ball reaches the ground after 2 seconds.

8. The speed-time graph for a car is shown in Fig. 8.12

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by car during the period. (b) Which part of the graph represents the uniform motion of the car?
Answer-
(a) 

Displacement of shaded area Is given as
= (1/2) X 4X6
= 12 meters.
the car travels a total of 12 meters in the first four seconds.

(b) The car is in uniform motion from the 6th to the 10th second.

9. State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero velocity (b) an object moving with an acceleration but with uniform speed. (c) an object moving in a certain direction with an acceleration in the perpendicular direction.

Answer-(a) if we have thrown a stone in opposite direction then it moves in an upward direction with uniform
acceleration due to gravity and on reaching. To a certain height, its velocity becomes zero and it starts to fall down. So this condition is possible.
(b) This condition is impossible.
(c) It is possible; if the object moves in a circular motion.

10. An artificial satellite is moving in a circular orbit of a radius of 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Answer-Radius = 42250 km
circumference of the orbit =2 π r
2 X 3.14 X42250km
= 265571.42 km
Time taken = 24 hours
, speed of the satellite = 265571.42/ 24
=11065.4 km/h.


 

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